I have a dataframe dbwith 2 categorical variables: varA has 4 levels (0,1,2,3), varB has 2 levels (yes,no). varB has no values for the level 0 of varA:
id varA varB
1 2 yes
2 3 no
3 3 no
4 1 yes
5 0 NA
6 1 no
7 2 no
8 3 yes
9 3 yes
10 2 no
I created a contingency table using CrossTable from the descr package and then a mosaic plot with the plot function:
table <- CrossTable(db$varA,db$varB, missing.include=FALSE)
plot(table,xlab="varA",ylab="varB")
I obtained this plot:
I would like to eliminate the level 0 from the plot. I also would like to add 2 y-axis, one on the left of the plot with a scale from 0 to 1 and one on the right with a scale from 1 to 0.
Could you help me?
Well, that was annoying. There is no support for subsetting such a "CrossTable" object. If it were a well-behaved table-like object you would been able to just pass table[ , -1] to the plot function. instead you need to do the subetting before the data that is passed to CrossTable:
table <- with( na.omit(db), CrossTable( varA, varB, missing.include=TRUE))
plot(table, xlab="varA", ylab="varB")
BTW using the name table for a data-object is quite confusing to regular R users since the table function is one of our basic tools.
Personally I would avoid avoid using that CrossTable function since its output is so weird and not available for management with typical R functions. Yeah, I know it produces a SAS-like output, but R users grow to love the compact output of the table function and the many matrix operations that are available for working with table-objects. You may need to get your margin percentages by hand with prop.table.
Related
I would like to add a column to my dataframe that contains categorical data based on numbers in another column. I found a similar question at Create categorical variable in R based on range, but the solution provided there didn't provide the solution that I need. Basically, I need a result like this:
x group
3 0-5
4 0-5
6 6-10
12 > 10
The solutions suggested using cut() and shingle(), and while those are useful for dividing the data based on ranges, they do not create the new categorical column that I need.
I have also tried using something like (please don't laugh)
data$group <- "0-5"==data[data$x>0 & data$x<5, ]
but that of course didn't work. Does anyone know how I might do this correctly?
Why didn't cut work? Did you not assign to a new column or something?
> data=data.frame(x=c(3,4,6,12))
> data$group = cut(data$x,c(0,5,10,15))
> data
x group
1 3 (0,5]
2 4 (0,5]
3 6 (5,10]
4 12 (10,15]
What you've created there is a factor object in a column of your data frame. The text displayed is the levels of the factor, and you can change them by assignment:
levels(data$group) = c("0-5","6-10",">10")
data
x group
1 3 0-5
2 4 0-5
3 6 6-10
4 12 >10
Read some basic R docs on factors and you'll get it.
I am a newbie in R. I need to generate some graphs. I imported an excel file and need to create a histogram on one column. My importing code is-
file=read.xlsx('femalecommentcount.xlsx',1,header=FALSE)
col=file[2]
col looks like this (part) -
36961 1
36962 1
36963 7
36964 1
36965 2
36966 1
36967 1
36968 4
36969 1
36970 6
36971 3
36972 1
36973 6
36974 6
36975 2
36976 2
36977 8
36978 2
36979 1
36980 1
36981 1
the first column is the row number. I'm not sure how to remove this. The second column is my data that I want a histogram on. hist() function requires a vector, I'm not sure how exactly to convert.
If I just simple call -
hist(col)
it gives-
Error in hist.default(col) : 'x' must be numeric
I have tried few commands randomly from the internet, but they didn't work.
My eventual goal is to just generate a good histogram (and maybe other charts) on that column, to get a good understadning of the spread of my data.
It should be col=file[[2]] or col=file[, 2] --- solution given in comment
data import should be in correct way to avoid numeric issue
I have data as follows in .csv format as I am new to ggplot2 graphs I am not able to do this
T L
141.5453333 1
148.7116667 1
154.7373333 1
228.2396667 1
148.4423333 1
131.3893333 1
139.2673333 1
140.5556667 2
143.719 2
214.3326667 2
134.4513333 3
169.309 8
161.1313333 4
I tried to plot a line graph using following graph
data<-read.csv("sample.csv",head=TRUE,sep=",")
ggplot(data,aes(T,L))+geom_line()]
but I got following image it is not I want
I want following image as follows
Can anybody help me?
You want to use a variable for the x-axis that has lots of duplicated values and expect the software to guess that the order you want those points plotted is given by the order they appear in the data set. This also means the values of the variable for the x-axis no longer correspond to the actual coordinates in the coordinate system you're plotting in, i.e., you want to map a value of "L=1" to different locations on the x-axis depending on where it appears in your data.
This type of fairly non-sensical thing does not work in ggplot2 out of the box. You have to define a separate variable that has a proper mapping to values on the x-axis ("id" in the code below) and then overwrite the labels with the values for "L".
The coe below shows you how to do this, but it seems like a different graphical display would probbaly be better suited for this kind of data.
data <- as.data.frame(matrix(scan(text="
141.5453333 1
148.7116667 1
154.7373333 1
228.2396667 1
148.4423333 1
131.3893333 1
139.2673333 1
140.5556667 2
143.719 2
214.3326667 2
134.4513333 3
169.309 8
161.1313333 4
"), ncol=2, byrow=TRUE))
names(data) <- c("T", "L")
data$id <- 1:nrow(data)
ggplot(data,aes(x=id, y=T))+geom_line() + xlab("L") +
scale_x_continuous(breaks=data$id, labels=data$L)
You have an error in your code, try this:
ggplot(data,aes(x=L, y=T))+geom_line()
Default arguments for aes are:
aes(x, y, ...)
Consider the following frequency data:
> table(income)
income
3 5 6 7 8 5000
2 7 2 2 2 1
When I type >hist(income) I get the following histogram
So as you can see, the fact that most income values are concentrated around 5 and there is one value quite distant from the others makes the histogram not look very good. MS Excel can consider the 5000 value as of another category, so the data would like this instead:
> table(income)
income
3 5 6 7 8 more
2 7 2 2 2 1
So plotting this as a histogram would look much better, so you can see the frequency within a shorter range:
Is there anyway to do this either with the hist() function or others functions from lattice or ggplot2? I do however, don't want to overwrite the values that exceed a certain threshold, so as I do lose any information.
Thanks a lot!
Data generation:
income <- c(rep(3,2), rep(5,7), rep(6,2), rep(7,2), rep(8,2), 5000)
Function for preparing data for plotting:
nice.data <- function(x, threshold=10){
x[x>threshold] <- "More"
x
}
Plotting:
library(ggplot2)
ggplot() + geom_histogram(aes(x=nice.data(income))) + xlab("Income")
Result:
Could you please help me to solve this problem:
I have a database like below:
Animal Milk Age
1 11.96703591 1
1 13.41236333 2
1 14.85769075 3
1 16.30301817 4
2 17.74834559 1
2 19.08465881 2
2 20.42097204 3
2 14.66094662 4
2 14.70197368 5
3 14.74300075 1
3 14.78402781 2
3 14.82505488 3
3 14.86608194 4
3 14.90710901 5
I want to make a plot between milk versus age, so I use function plot(Milk~Age, data=mydata)
My question is how can I make the same plot (Milk~Age) for each individual, by using only one function. Since I have about 200 animals and if I have to run 200 times to produce 200 curves.
Thanks
Phuong
One approach would be to use library ggplot2 and then make individual facets for each animal. As you have many animals you can change ncol= or nrow= in facet_wrap() to get better view.
library(ggplot2)
ggplot(df,aes(x=Age,y=Milk))+geom_point()+facet_wrap(~Animal)
The following code should create as many plot as you have unique Animal values, and store them in different pdf files in the working directory :
invisible(by(df, df$Animal, function(tmpdf) {
pdf(paste0("plot",tmpdf$Animal[1],".pdf"))
plot(Milk~Age, data=tmpdf, main=tmpdf$Animal[1])
dev.off()
}))
I would say to use ggplot from the ggplot2 package
ggplot(df,aes(x=Age,y=Milk, color=Animal))+geom_point()
edit1: actually this would lose clarity with 200 animals. Did you want all this data point in one graph or spread out across 200 graphs? If the latter then I agree with Didzis